Question

A convex lens of focal length $15\text{cm}$ is split into two halves and the two halves are placed at a separation of $120\text{cm}$. Between the two halves of convex lens a plane mirror is placed horizontally and at a distance of $4\text{mm}$ below the principal axis of the lens halves. An object of length $2\text{mm}$ is placed at a distance of $20\text{cm}$ from one half lens as shown in figure.


Find the position and size of the final image.

• A. $20\text{cm}$ right of $L_2,2\text{mm}$ length
• B. $10\text{cm}$ right of $L_2,2\text{mm}$ length
• C. $20\text{cm}$ left of $L_2,1\text{mm}$ length
• D. None of the above

Answer 1

The correct option is A $20\text{cm}$ right of $L_2,2\text{mm}$ length

For refraction at first half lens $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{-20}=\frac{1}{15}$
$\therefore v=60\text{cm}$
Magnification , $m=\frac{v}{u}=\frac{60}{-20}=-3$
The image formed by first half lens is shown in figure (a).
Here $AB=2\text{mm},A_1{B}_1=6\text{mm},A{O}_1=20\text{cm}{O}_{1}F=15\text{cm}{O}_1{A}_1=60\text{cm}$.

Point $B_1$ is $6\text{mm}$ below the principal axis of the lenses. Plane mirror is $4\text{mm}$ below it.
Hence, $4\text{mm}$ length of $A_1{B}_1$ (i.e. $A_1{C}_1)$ acts as real object for mirror. Mirror forms its virtual image $A_2{C}_2.2\text{mm}$ length of $A_1{B}_1$ (i.e. $C_1{B}_1)$ acts as virtual object for mirror. Real image $C_2{B}_2$ is formed of this part. Image formed by plane mirror is shown in figure (b).

Here $A_2{B}_2=6\text{cm}$

For the second half lens,

$\frac{1}{v}-\frac{1}{-60}=\frac{1}{15}$
$\therefore v=+20$
Magnification is
$m=\frac{v}{u}=\frac{20}{-60}=-\frac{1}{3}$
So, length of final image
$A_3{B}_3=\frac{1}{3}{A}_2{B}_2=2\text{mm}$